# Spring 2016, problem 8

Find all continuous functions $f : \mathbf{R}\to\mathbf{R}$ such that $$f (f (x)) = f (x)+x,$$ for any $x\in \mathbf{R}$.

Deca
5 years ago

Hi. I am Debinson from Bogotá-Colombia f(f(x))=f(x)+x

Previous ideas: We find the solution of the next equation: f(f(x))=f(x) In this case we would have that f(x) must be a constant function or identity function. We find the solution of the next equation:

f(f(x))=x

In this case we would have that f(x) must be identity function. Thus the function f(x) must be of the form f(x)=ax and we would have: f(f(x))=f(x)+x a^2 x=ax+x Which bring us: x∙(a^2-a-1)=0 And we get: a=(1±√5)/2 Requested functions are: f(x)=(1±√5)/2 x

I understand your solution but I was having a problem that you did X=10Y+R, why you have taken 10 in particular, I mean you can take any other integer like 9,11 etc.

countess179 1 month ago
krishnaraj
6 years ago

f(x) has to be a linear function, or else f(f(x)) cannot be of the same order as f(x) + x which is required by the given equation. Once we observe this, it's straightforward what this linear function is. f(x) = [(1+/-sqrt(5))/2]x.

Hubert
6 years ago

Hi, I'm Hubert from Paris ...

• $f$ is injective by $g_x=f_x-x\Rightarrow~g(f_x)=x$, and so continuously monotonic, $\lim_{+\infty} f=\alpha$ contradicts $(*)$ hence $|\lim_{+\infty}f|=\infty$, by same way $|\lim_{-\infty}f|=\infty$; hence $f$ is surjective and one to one, as $\phi=f^{-1}$,

$(*) f(f_x)=f_x+x\Rightarrow~\phi^2_x+\phi_x=x$

$x_n=\phi^n(x)=\phi(\phi(...(x)))\Rightarrow~x_{n+2}+x_{n+1}=x_n\Rightarrow~x_n=Aa^n+B.b^n$

with $-b=\frac 1a=\varphi=\frac{\sqrt 5+1}2$, solving yields $A\sqrt 5=bx-\phi_x; B\sqrt 5=\phi_x-ax$;

$f_0=0$ by $f_0>0\Rightarrow~0+f_0=f^2_0>f_0$: contradiction; same way for $f_0 \< 0$, $x>0\Rightarrow~x-\phi_x=\phi^2_x>0\Rightarrow~\phi_x\<x\rightarrow~>$B\sqrt 5=\phi_x-ax=0 ~(|b|>1\Rightarrow |b^n|\longrightarrow\infty)\Rightarrow~f_x=\frac xa=\varphi.x$which works, same way for$x\<0~$, EDITED - Right ...this is only the increasing solution, same idea for the decreasing one ... Nelix 6 years ago Two such functions are given by$ f(x):= ax$where either$ a=\phi$or$ a=- \frac{1}{\phi}$with$\phi$the golden ratio. But I don't think those are the only examples. edit: I was mistaken. These are the only examples. Heres a proof, similiar, i guess, to huberts, which, I admit, I find impossible to follow (and something's missing there, anyway, because$f(x) = \phi x$is not the only solution). Consider the basic equation$f(f(x)) - f(x) = x$. Since the right hand side depends on$x$, but the LHS only depends on$f(x)$,$f$has to be injective and therefore strictly monotonous. Furthermore, as hubert correctly states,$|\lim_{x \to \infty} f(x)| \lt \infty$or$| \lim_{x \to -\infty} f(x)| \lt \infty$yields an immediate contradiction. Thus$f$is surjective. The inverse$f^{-1}$therefore exists. For natural$n$define as usual •$f^n$is f applied$n$times and •$f^{-n}$is$f^{-1}$applied$n$times. •$f^0(x) := x$For given$x$, the (two-sided) sequence$..., f^{-2}(x), f^{-1}(x), x, f^{1}(x), f^{2}(x), ...$is the (two-sided) Fibonacci sequence for initial values$x$and$f(x)$. The solution is given by: $$f^z(x) = u \phi^z + v (-\phi)^{-z}$$ with$u$and$v$constants. Now pick an arbitrary nonzero$a_0 \in \mathbb{R} \setminus \left \{ 0 \right \}$. Note that$f(a_0) \neq a_0$, as the contrary contradicts the basic equation. Define$a_z := f^z(a_0) =: u_a \phi^z + v_a (-\phi)^{-z}$. We cannot, without loss of generality assume, that$a_1 > a_0$, but it turns out, that no new arguments are involved for the opposite case, so let's assume that, anyway. Now we discern two cases: Case 1 ($f$is strictly increasing): In this case$\forall z \in \mathbb{Z}: a_{z} \lt a_{z + 1} $. But if$v_a$nonzero and we let$z \rightarrow - \infty$the term$v_a (-\phi)^{-z}$will dominate, turning that inequality around every other time. Thus$v_a = 0$and$f(a) = \phi a$and we're finished. Case 2 ($f$is strictly decreasing): In this case$a_z \lt a_{z+1}$if z is even and$a_z > a_{z+1}$otherwise. But if$u_a \neq 0$and we let$z \rightarrow + \infty$, the term$u_a \phi^z$will dominate, so that only one of the equalities can be satisfied for large$z$. Thus$u_a = 0$and$f(a) = -\frac{1}{\phi} a \$ and we're finished.